Promise to brianliest!
Find every value of the variable that makes each rational expression undefined.
Read the picture! 1-5 only
Thank you in advance!
nonsense/sp3m answers = report
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Promise to brianliest!
Find every value of the variable that makes each rational expression undefined.
Read the picture! 1-5 only
Thank you in advance!
nonsense/sp3m answers = report
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Answer:
Certainly! Let's go through each expression one by one and explain the values of the variable that make them undefined.
1. \( \frac{3x}{x-5} \):
In this expression, the denominator is \( x-5 \). For the expression to be undefined, the denominator must be equal to zero. So, we solve the equation \( x-5 = 0 \) and find that \( x = 5 \). Therefore, when \( x = 5 \), the expression becomes undefined.
2. \( \frac{2x+5}{x^2+11x+10} \):
Here, the denominator is \( x^2+11x+10 \). To make the expression undefined, we need to find the values of \( x \) that make the denominator zero. By factoring the quadratic equation \( x^2+11x+10 = 0 \), we get \( (x+1)(x+10) = 0 \). Solving for \( x \), we find that \( x = -1 \) and \( x = -10 \). Therefore, when \( x = -1 \) or \( x = -10 \), the expression becomes undefined.
3. \( \frac{5r}{r^2-16} \):
In this expression, the denominator is \( r^2-16 \). To make the expression undefined, we set the denominator equal to zero and solve the equation \( r^2-16 = 0 \). By factoring, we get \( (r-4)(r+4) = 0 \). Solving for \( r \), we find that \( r = -4 \) and \( r = 4 \). Therefore, when \( r = -4 \) or \( r = 4 \), the expression becomes undefined.
4. \( \frac{4y}{y+8} \):
Here, the denominator is \( y+8 \). To make the expression undefined, we need to find the values of \( y \) that make the denominator zero. Solving the equation \( y+8 = 0 \), we find that \( y = -8 \). Therefore, when \( y = -8 \), the expression becomes undefined.
5. \( \frac{7}{x^2-2x-15} \):
In this expression, the denominator is \( x^2-2x-15 \). To make the expression undefined, we set the denominator equal to zero and solve the equation \( x^2-2x-15 = 0 \). By factoring, we get \( (x-5)(x+3) = 0 \). Solving for \( x \), we find that \( x = 5 \) and \( x = -3 \). Therefore, when \( x = 5 \) or \( x = -3 \), the expression becomes undefined.
6. \( \frac{3x}{x^2-16x+15} \):
Here, the denominator is \( x^2-16x+15 \). To make the expression undefined, we need to find the values of \( x \) that make the denominator zero. Solving the equation \( x^2-16x+15 = 0 \), we find that \( x = 1 \) and \( x = 15 \). Therefore, when \( x = 1 \) or \( x = 15 \), the expression becomes undefined.
In summary, the values of the variable that make each expression undefined are:
1. \( x = 5 \)
2. \( x = -1 \) and \( x = -10 \)
3. \( r = -4 \) and \( r = 4 \)
4. \( y = -8 \)
5. \( x = 5 \) and \( x = -3 \)
6. \( x = 1 \) and \( x = 15 \)